finite dimensional ideal of properly infinite von Neumann algebra is $\{0\}$.

Question

Let $J$ be finite dimensional two sided ideal of a properly infinite von Neumann algebra $M$. How would one show that $J=\{0\}$?

Answer 1

If $J$ is finite-dimensional, it is closed, and it is then a finite-dimensional C$^*$-algebra. If it is not zero, it contains a nonzero projection $P$.

As $M$ is properly infinite, it has a pair of isometries $S,T$ with $S^*S=T^*T=I$ and $SS^*+TT^*=I$ (by the Halving Theorem, applied to $I$).

We have that $SPS^*$ and $TPT^*$ are nonzero pairwise orthogonal projections in $J$, since $S^*T=0$. Then $$ S^2P(S^*)^2,\ STPT^*S^*,\ TPT^*$$ are three nonzero pairwise orthogonal projections in $J$.

We can continue, inductively, to show that if $Q_1,\ldots,Q_m$ are pairwise orthogonal projections in $J$, then $$ SQ_1S^*,\ldots,SQ_mS^*, TPT^* $$ are $m+1$ nonzero pairwise orthogonal projections in $J$. It follows that $J$ contains arbitrarily large chains of pairwise orthogonal projections; so it is infinite-dimensional, a contradiction.